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A 1,160 kg satellite orbits earth with a tangential speed of 7,446 m/s. if the satellite experiences a centripetal force of 8,955 n, what is the height of the satellite above the surface of earth? recall that earth’s radius is 6.38 × 106 m and earth’s mass is 5.97 × 1024 kg. 3.71 × 1028 m 8.02 × 105 m 7.20 × 105 m 9.67 × 1028 m

Sagot :

The satellite is 8.02 × 10⁵ m above Earth's surface.

Let H be the height above the surface of the Earth; since we know that the satellite is rotating around the Earth due to the gravitational pull of the planet, we may assert

Procedure to solve:

F = mv²/R+H

H = mv²/F - R

H = (1160 × 7446²/8955 - 6.38 × 10⁶)

M = 8.02 × 10⁵ m

About centripetal force:

The force applied to an item that is in velocity of  curved motion that is pointed toward the axis of rotation or the centre of curvature is known as a centripetal force.

The centripetal force formula is given as the product of mass (in kg) and tangential velocity (in meters per second) squared, divided by the radius (in meters) that implies that on doubling the tangential velocity, the centripetal force will be quadrupled. Mathematically it is written as:

F = mv²/r

Learn more about velocity here:

https://brainly.com/question/18084516

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revakolhe

Answer:

8.02x 10^5 m

Explanation:

quiz

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